A bucket is being lifted from the bottom of a 50-foot deep well; its weight (including the water), \(B\text{,}\) in pounds at a height \(h\) feet above the water is given by the function \(B(h)\text{.}\) When the bucket leaves the water, the bucket and water together weigh \(B(0) = 20\) pounds, and when the bucket reaches the top of the well, \(B(50) = 12\) pounds. Assume that the bucket loses water at a constant rate (as a function of height, \(h\)) throughout its journey from the bottom to the top of the well.

*(a)* Find a formula for \(B(h) =\)

*(b)* Compute the value of the product \(B(5) \triangle h\text{,}\) where \(\triangle h = 2\) feet. Notice that this product represents the approximate work it took to move the bucket of water from \(h = 5\) to \(h = 7\text{.}\)

\(B(5) \triangle h =\)

What are the units on the product

\(B(5) \triangle h =\text{?}\)
lbs

ft

lbs/ft

ft/lbs

ft*ft

lbs*lbs

lbs*ft

*(c)* Is the value in (b) an over- or under-estimate of the actual amount of work it took to move the bucket from \(h = 5\) to \(h = 7\text{?}\) Think about why your answer is true.

Answer:

Overestimate

Underestimate

*(d)* Compute the value of the product \(B(22) \triangle h\text{,}\) where \(\triangle h = 0.25\) feet.

\(B(22) \triangle h =\)

What are the units on the product

\(B(22) \triangle h =\text{?}\)
lbs

ft

lbs/ft

ft/lbs

ft*ft

lbs*lbs

lbs*ft

*(e)* Notice that the value in part (d) estimates the amount of work it takes to move the bucket from a height of 22 feet to 22.25 feet. More generally, the quantity \(W_{\small\mbox{slice}} = B(h) \triangle h\) measures the amount of work it takes to move from a given value of \(h\) to a small positive value \(\triangle h\text{.}\)

*(f)* Evaluate the definite integral \(\int_0^{50} B(h) \, dh =\) .

What is the meaning of the value you find? Why?